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I learned that the wavefunction for the hydrogen atom can be solved analytically (we did the derivation in class), but that for more complicated atoms it is "impossible" to solve and that only approximations can be obtained.

Well, I'm not sure I like the vague wording with "impossible".

Basically, a hydrogen atom can be solved in terms of elementary functions. Elementary functions, as far as I know, are just arbitrarily defined (Is $\sqrt{}$ one? What about $\text{erfc}$?). For instance, I could say that $\sin{x}$ cannot be solved exactly except in a few special cases (x = some rational multiples of $\pi$).

To analogize, a general quintic equation cannot be "solved" exactly using elementary functions -- that is, unless you define a Bring radical as a new elementary function. To me, it's really arbitrary and all of these solutions come down to numerical, iterative refinement. Solved exactly vs solved approximately seems to be an ill-defined concept.

So what I am asking -- given enough computer time -- can I solve the Schrodinger equation of, say, Lithium, to whatever precision I like? Much as how I can solve $\sin{x}$ to however many decimal places I want? Or is there some other limit to such a calculation?

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3 Answers 3

up vote 8 down vote accepted

You are right about exact results, these depend on your definition of "exact". The best definition of an exact is if you have a fast algorithm to calculate the result in a reasonable time. The faster the algorithm, the more exact the result.

For Helium atoms, the answer is yes--- you can use the variational method to produce a result to as good a precision as you like, in short time. You can even do the first approximations with pencil and paper. Except for the pencil and paper, this is also true for Li, Be, B, C, maybe as high as Ne.

But for atoms like Iron, you get another problem which is that the space of all wavefunctions is growing exponentially. This means that describing the energy levels of the atom with all its interparticle entanglements in some highly excited state, is very very difficult. It cannot be solved in general, because solving the general quantum system includes quantum computation, which can solve some things, like factoring integers, which are very unlikely to have an efficient classical algorithm.

But for calculating the ground state wavefunction of Fermions, and levels near the ground state, you have two methods that are practical:

  • Hartree Fock--- this assumes the electrons move in a potential which is the nuclear potential, and a central potential due to the other electrons. This is a self-consistent method, and is good enough for practical atomic physics.
  • Density functional theory--- here you try to reparametrize the ground state wavefunction in terms of a three dimensional function which represents the electron density, and use tricks to figure out the details of the electronic entanglements inasmuch as these are necessary.

Both these methods are best for ground states and near-ground state properties. In general, the resources you need grow exponentially in the number of electrons, but these methods are practical for nearly all atoms and molecules you will ever encounter, and they can be refined by higher order approximations (although this is more practical for DFT).

The bosonic ground state problem is very simple in comparison to the electron problem you are asking about, and requires only monte-carlo on the imaginary time version. This is how quantum fields are simulated.

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Let me take a slightly different perspective to Ron Maimon and say that the answer depends on whether you're after an exact solution of some mathematical model, or whether you want to calculate the exact physical behaviour.

Any method for calculating the physical properties of a system rely upon an approximation. If you choose some model you can certainly (in principle) do your calculation to arbitrary accuracy, but this is an exact calculation of the model and not of reality. Notoriously, the most exact calculation we've done is of the electron gyromagnetic ratio, $g$. However it's believed (not proven as far as I know) that the perturbation methods used diverge eventually, so even an arbitrarily powerful computer cannot calculate $g$ to any accuracy. To improve the accuracy of the calculation would require an improved mathematical model of the physics, and as far as I know such a model does not exist at present.

But I should emphasise that although I would say the answer to your question is that no we can't calculate to arbitrary precision, this is a somewhat pedantic point of view since we can in principle calculate to better precision than we're ever likely to measure.

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Your original question is ambiguous. Numerical methods have memory requirements as well. If the memory in your computer is not enough to store the solution then it does not matter how many time you wait for the numerical solution: The algorithm will break with some out of memory error.

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That's not the problem with QM--- memory is not an issue, time to run is. Complexity theorists say this by saying BQP is inside PSPACE, although not inside P, meaning you can simulate quantum mechanics with a polynomial amount of memory, but an exponential number of steps. –  Ron Maimon Oct 13 '12 at 21:23
    
Ron, out-of-memory errors are rather usual in quantum chemical software even for small systems and not too large precision runs.There is not enough memory in our planet for arbitrary precision computations in large quantum systems. –  juanrga Oct 15 '12 at 9:49
    
There was problem with above comment: @RonMaimon, out-of-memory errors are rather usual in quantum chemical software even for small systems and not too high precision runs. See Biowulf and Chemistry High Performance Computing. There is not enough memory in our planet for arbitrary precision computations in large quantum systems, even if you have the eternity to wait you will never see the solution because the algorithm breaks. –  juanrga Oct 15 '12 at 9:59
    
This is probably true in practice, but the amount of memory required only scales linearly with the size of the system for DFT, I don't think you would need to much memory nor too much running time for Car-Parinello method, which is not arbitrary precision, since it uses Hartree Fock on inner shell electrons, but good enough for all usual chemistry. –  Ron Maimon Oct 15 '12 at 15:11
    
@RonMaimon: Today all competitive SCF, DFT, and MP2 programs commonly use Integral-Direct methods for large basis sets (those methods were developed to overcome the integral-storage bottleneck). As explained here "The wavefunction memory size is usually the limiting factor in Car-Parrinello calculations". If you choose a large enough quantum system then, as said before "There is not enough memory in our planet for arbitrary precision computations" and eternity is not the problem. –  juanrga Oct 16 '12 at 10:47

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